3.
4 pN = 8.8 × 10
-13
pounds. The weight of a single motor domain is 100,000 g mol
-1
(6.023 × 10
23
molecules mol
-1
) =
1.7 × 10
-19
g = 7.6 × 10
-23
pounds. Thus, a motor domain can lift (8.8 × 10
-13
/7.6 × 10
-23
) = 1.2 × 10
10
times its
weight.
See question
4.
After death, the ratio of ADP to ATP increases rapidly. In the ADP form, myosin motor domains bind tightly to
actin. Myosin-actin interactions are possible because the drop in ATP concentration also allows the calcium
concentration to rise, clearing the blockage of actin by tropomyosin through the action of the troponin complex.
See question
5.
Above its critical concentration, ATP-actin will polymerize. The ATP will hydrolyze through time to form ADP-
actin, which has a higher critical concentration. Thus, if the initial subunit concentration is between the critical
concentrations of ATP-actin and ADP-actin, filaments will form initially and then disappear on ATP hydrolysis.
See question
6.
Actin monomers bind and hydrolyze ATP and are in somewhat different conformations, depending on the identity of
the bound nucleotide. Thus, nucleotide binding and then hydrolysis by actin filaments might change actin's length,
shape, or rigidity.
See question
7.
The first (a) should behave like conventional kinesin and move toward the plus end of microtubules, whereas the
second (b) should behave like ncd and move toward the minus end.
See question
8.
A one-base step is approximately 3.4 Å = 3.4 × 10
-4
µm. If a stoichiometry of one molecule of ATP per step is
assumed, this distance corresponds to a velocity of 0.017 µm s
-1
. Kinesin moves at a velocity of 6400 Å per second,
or 0.64 µm s
-1
.
See question
9.
A protonmotive force across the plasma membrane is necessary to drive the flagellar motor. Under conditions of
starvation, this protonmotive force is depleted. In acidic solution, the pH difference across the membrane is
sufficient to power the motor.
See question
10.
(a) The force is 6π(0.01 g cm
-1
s
-1
)(0.0002 cm)(0.00006 cm s
-1
) = 2.3 × 10
-9
dyne. (b) The work performed is (2.3
× 10
-9
dyne)(0.00006 cm) = 1.4 × 10
-13
erg. (c) On the basis of a ∆ G value of -12 kcal mol
-1
(50 kJ mol
-1
) under
typical cellular conditions, the energy available is 8.3 × 10
-13
erg per molecule. In 1 second, 80 molecules of ATP
are hydrolyzed, corresponding to 6.6 × 10
-11
erg. Thus, a single kinesin motor provides more than enough free
energy to power the transport of micrometer-size cargoes at micrometer-per-second velocities.
See question